The value of power dissipated across the zener diode (Vz = 15V) connected in the circuit as shown in the figure is x × 10 - 1 watt. The value of x, to the nearest integer, is __________. Numerical answer type. Work it out, then open the answer.

Answer: 5. Without the Zener the load gets 22 × 90/125 = 15.84 V > 15 V, so the Zener holds 15 V. Is = (22 - 15)/(35) = 0.2 A, IL = 15/90 = 1/6 A IZ = 0.2 - 1/6 = 1/30 A P = 15 × 1/30 = 0.5 W = 5 × 10⁻¹ W, so x = 5.

The value of power dissipated across the zener diode (Vz = 15V)… | JEE Main 2021 (16 Mar, Shift 1)

Q70JEE Main 2021 · 16 MAR · SHIFT 1Medium

Special Purpose Diodes

The value of power dissipated across the zener diode (Vz = 15V) connected in the circuit as shown in the figure is x $\times$ 10 $-$ 1 watt.

The value of x, to the nearest integer, is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Without the Zener the load gets $22 \times \frac{90}{125} = 15.84$ V $> 15$ V, so the Zener holds $15$ V.
$I_s = \frac{22 - 15}{35} = 0.2$ A, $I_L = \frac{15}{90} = \frac{1}{6}$ A
$I_Z = 0.2 - \frac{1}{6} = \frac{1}{30}$ A
$P = 15 \times \frac{1}{30} = 0.5$ W $= 5 \times 10^{-1}$ W, so $x = 5$ .

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